bpf: change eBPF helper doc parsing script to allow for smaller indent

Documentation for eBPF helpers can be parsed from bpf.h and eventually
turned into a man page. Commit 6f96674dbd ("bpf: relax constraints on
formatting for eBPF helper documentation") changed the script used to
parse it, in order to allow for different indent style and to ease the
work for writing documentation for future helpers.

The script currently considers that the first tab can be replaced by 6
to 8 spaces. But the documentation for bpf_fib_lookup() uses a mix of
tabs (for the "Description" part) and of spaces ("Return" part), and
only has 5 space long indent for the latter.

We probably do not want to change the values accepted by the script each
time a new helper gets a new indent style. However, it is worth noting
that with those 5 spaces, the "Description" and "Return" part *look*
aligned in the generated patch and in `git show`, so it is likely other
helper authors will use the same length. Therefore, allow for helper
documentation to use 5 spaces only for the first indent level.

Signed-off-by: Quentin Monnet <quentin.monnet@netronome.com>
Signed-off-by: Daniel Borkmann <daniel@iogearbox.net>
This commit is contained in:
Quentin Monnet 2018-05-17 13:43:56 +01:00 committed by Daniel Borkmann
parent b9f672af14
commit eeacb7166d
1 changed files with 4 additions and 4 deletions

View File

@ -95,7 +95,7 @@ class HeaderParser(object):
return capture.group(1)
def parse_desc(self):
p = re.compile(' \* ?(?:\t| {6,8})Description$')
p = re.compile(' \* ?(?:\t| {5,8})Description$')
capture = p.match(self.line)
if not capture:
# Helper can have empty description and we might be parsing another
@ -109,7 +109,7 @@ class HeaderParser(object):
if self.line == ' *\n':
desc += '\n'
else:
p = re.compile(' \* ?(?:\t| {6,8})(?:\t| {8})(.*)')
p = re.compile(' \* ?(?:\t| {5,8})(?:\t| {8})(.*)')
capture = p.match(self.line)
if capture:
desc += capture.group(1) + '\n'
@ -118,7 +118,7 @@ class HeaderParser(object):
return desc
def parse_ret(self):
p = re.compile(' \* ?(?:\t| {6,8})Return$')
p = re.compile(' \* ?(?:\t| {5,8})Return$')
capture = p.match(self.line)
if not capture:
# Helper can have empty retval and we might be parsing another
@ -132,7 +132,7 @@ class HeaderParser(object):
if self.line == ' *\n':
ret += '\n'
else:
p = re.compile(' \* ?(?:\t| {6,8})(?:\t| {8})(.*)')
p = re.compile(' \* ?(?:\t| {5,8})(?:\t| {8})(.*)')
capture = p.match(self.line)
if capture:
ret += capture.group(1) + '\n'